Classical: inductor current, 2 V charged 1 F capacitor

Circuit reading. After the switch closes, L, R = 10 Ω and C = 1 F (v_C(0) = 2 V) are connected across the same two nodes, so it is a source-free parallel RLC circuit. Initially the switch is open, so i_L(0) = 0. Take v as the voltage across the capacitor (+ on the left plate) and i_L as the inductor current driven by it.

Equations

C dv/dt + v/R + i_L = 0
L di_L/dt = v
→ d²i_L/dt² + (1/RC) di_L/dt + (1/LC) i_L = 0
→ i_L'' + 0.1 i_L' + 2 i_L = 0

Roots

m² + 0.1m + 2 = 0 → m = −0.05 ± j√(2 − 0.0025) = −0.05 ± j1.4133

α = 1/(2RC) = 0.05 and ω₀ = 1/√(LC) = √2 = 1.414: underdamped (α < ω₀), ω_d = 1.4133 rad/s.

Constants

• i_L(0) = 0 → i_L = e^(−0.05t) (A cos ω_d t + B sin ω_d t) with A = 0
• i_L′(0) = v(0)/L = 2/0.5 = 4 A/s. With A = 0: B ω_d = 4 → B = 4/1.4133 = 2.830

Answer: i_L(t) = 2.83 e^(−0.05t) sin(1.4133t) A, t > 0 (opposite capacitor polarity only reverses the sign).

The current is a slowly decaying oscillation (period 2π/ω_d = 4.45 s). Check: at t = 4 s the formula gives −1.365 A, the same as a numerical integration of the circuit equations; i_L(0) = 0 and i_L′(0) = 2.83 × 1.4133 = 4 A/s = v(0)/L ✓.